(0) Obligation:
Clauses:
ordered([]).
ordered(.(X, [])).
ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs))).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(0)).
le(0, 0).
Query: ordered(g)
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
ordered_in: (b)
le_in: (b,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
ORDERED_IN_G(
x1) =
ORDERED_IN_G(
x1)
U1_G(
x1,
x2,
x3,
x4) =
U1_G(
x2,
x3,
x4)
LE_IN_GG(
x1,
x2) =
LE_IN_GG(
x1,
x2)
U3_GG(
x1,
x2,
x3) =
U3_GG(
x3)
U2_G(
x1,
x2,
x3,
x4) =
U2_G(
x4)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
ORDERED_IN_G(
x1) =
ORDERED_IN_G(
x1)
U1_G(
x1,
x2,
x3,
x4) =
U1_G(
x2,
x3,
x4)
LE_IN_GG(
x1,
x2) =
LE_IN_GG(
x1,
x2)
U3_GG(
x1,
x2,
x3) =
U3_GG(
x3)
U2_G(
x1,
x2,
x3,
x4) =
U2_G(
x4)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
LE_IN_GG(
x1,
x2) =
LE_IN_GG(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(8) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(9) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (EQUIVALENT transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
(13) YES
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
The TRS R consists of the following rules:
ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, Y)) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))
The argument filtering Pi contains the following mapping:
ordered_in_g(
x1) =
ordered_in_g(
x1)
[] =
[]
ordered_out_g(
x1) =
ordered_out_g
.(
x1,
x2) =
.(
x1,
x2)
U1_g(
x1,
x2,
x3,
x4) =
U1_g(
x2,
x3,
x4)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
U2_g(
x1,
x2,
x3,
x4) =
U2_g(
x4)
ORDERED_IN_G(
x1) =
ORDERED_IN_G(
x1)
U1_G(
x1,
x2,
x3,
x4) =
U1_G(
x2,
x3,
x4)
We have to consider all (P,R,Pi)-chains
(15) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
The TRS R consists of the following rules:
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
le_in_gg(
x1,
x2) =
le_in_gg(
x1,
x2)
s(
x1) =
s(
x1)
U3_gg(
x1,
x2,
x3) =
U3_gg(
x3)
0 =
0
le_out_gg(
x1,
x2) =
le_out_gg
ORDERED_IN_G(
x1) =
ORDERED_IN_G(
x1)
U1_G(
x1,
x2,
x3,
x4) =
U1_G(
x2,
x3,
x4)
We have to consider all (P,R,Pi)-chains
(17) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U1_G(Y, Xs, le_out_gg) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, le_in_gg(X, Y))
The TRS R consists of the following rules:
le_in_gg(s(X), s(Y)) → U3_gg(le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg
le_in_gg(0, 0) → le_out_gg
U3_gg(le_out_gg) → le_out_gg
The set Q consists of the following terms:
le_in_gg(x0, x1)
U3_gg(x0)
We have to consider all (P,Q,R)-chains.
(19) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
le_in_gg(s(X), s(Y)) → U3_gg(le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg
le_in_gg(0, 0) → le_out_gg
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 2·x1 + 2·x2
POL(0) = 0
POL(ORDERED_IN_G(x1)) = x1
POL(U1_G(x1, x2, x3)) = 2·x1 + 2·x2 + 2·x3
POL(U3_gg(x1)) = x1
POL(le_in_gg(x1, x2)) = x1 + x2
POL(le_out_gg) = 0
POL(s(x1)) = x1
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U1_G(Y, Xs, le_out_gg) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, le_in_gg(X, Y))
The TRS R consists of the following rules:
U3_gg(le_out_gg) → le_out_gg
The set Q consists of the following terms:
le_in_gg(x0, x1)
U3_gg(x0)
We have to consider all (P,Q,R)-chains.
(21) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(22) TRUE